Question

A cylinder with a piston contains 0.250 mol of oxygen at 2.40 \(\times\) 10\(^5\) Pa and 355 K. The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(pV\)-diagram. Compute (b) the temperature during the isothermal compression; (c) the maximum pressure; (d) the total work done by the piston on the gas during the series of processes.

Short answer

The processes form a cycle on a pV-diagram. The temperature is 355 K during isothermal compression, maximum pressure is approximately 4.80 × 10^5 Pa, and total work done is 227 J.

Step-by-step solution

Determine Initial Conditions

First, calculate the initial volume of the gas using the ideal gas law, \(PV = nRT\). We know: \(P = 2.40 \times 10^5\,\mathrm{Pa}\), \(n = 0.250\,\mathrm{mol}\), \(R = 8.314\,\mathrm{J/(mol\cdot K)}\), and \(T = 355\,\mathrm{K}\). Therefore, \(V = \frac{nRT}{P}\). Substitute the values to find \(V\).\[\begin{align*} V &= \frac{0.250 \times 8.314 \times 355}{2.40 \times 10^5} \ &= 3.074 \times 10^{-3}\,\mathrm{m}^3. \end{align*}\]

Construct pV Diagram

In an isobaric expansion, the pressure remains constant while the volume doubles. Starting at (\(V=3.074 \times 10^{-3}\,\mathrm{m}^3\), \(P = 2.40 \times 10^5\,\mathrm{Pa}\)) and moving horizontally on the \(pV\)-diagram to (\(V=2\times 3.074 \times 10^{-3}\,\mathrm{m}^3\)). Next, an isothermal process follows, bringing the volume back to \(3.074 \times 10^{-3}\,\mathrm{m}^3\) as the pressure changes. Finally, an isochoric process occurs from this point, returning the system to its initial pressure.

Calculate Temperature during Isothermal Compression

As the gas is compressed isothermally back to its original volume, the temperature remains constant. Therefore, the temperature during this process is \(355\,\mathrm{K}\).

Determine Maximum Pressure

To find the maximum pressure during the compression, use the ideal gas law at the end of the isothermal compression. Since the volume returns to \(3.074 \times 10^{-3}\,\mathrm{m}^3\), the new pressure \(P'\) is given by \(P'V=nRT\). Thus, substitute \(V=3.074 \times 10^{-3}\,\mathrm{m}^3\), \(T=355\,\mathrm{K}\) and solve for \(P'\).\[\begin{align*} P' &= \frac{nRT}{V} \ &= \frac{0.250 \times 8.314 \times 355}{3.074 \times 10^{-3}} \approx 4.80 \times 10^5 \, \mathrm{Pa}. \end{align*}\]

Calculate Total Work Done

The work done in an isobaric process is given by \(W_1 = P\Delta V\) and during isothermal compression by \(W_2 = nRT \ln(\frac{V_f}{V_i})\). The isochoric process does no work. Calculate \(W_1\) with the initial volume being \(3.074 \times 10^{-3}\,\mathrm{m}^3\) and \(\Delta V\), followed by \(W_2\) using the final volume back to the initial \(3.074 \times 10^{-3}\,\mathrm{m}^3\). \[\begin{align*} W_1 &= 2.40\times 10^5 \times (2\cdot3.074\times 10^{-3} - 3.074\times 10^{-3}) \ &= 739.2\,\mathrm{J},\W_2 &= 0.250 \times 8.314 \times 355 \times \ln \left(\frac{3.074 \times 10^{-3}}{2 \times 3.074 \times 10^{-3}}\right)\&=-512.2\,\mathrm{J}.\end{align*}\] The total work done is \(W_1 + W_2 = 739.2 - 512.2 = 227\,\mathrm{J}.\)

Key concepts

pV diagrams

PV diagrams, or pressure-volume diagrams, are a useful way to visualize the changes that occur in a gas due to different processes. They provide a graphic representation of a process by plotting pressure (P) on the y-axis and volume (V) on the x-axis. This type of diagram helps us understand the behavior of gases in accordance with the ideal gas laws. The area under a curve on a PV diagram represents the work done in the process if the units are in standard pressure and volume units.

In the given exercise, the process involves three key stages: an isobaric expansion, an isothermal compression, and an isochoric cooling. These transformations can be illustrated on the PV diagram:

• The isobaric process is shown as a horizontal line on the PV diagram.
• The isothermal process appears as a curved line returning to the initial volume but with a changed pressure.
• The isochoric process is represented by a vertical line at constant volume.

This visual representation helps in quickly understanding the nature of the thermodynamic path the gas undergoes, and therefore, it's a critical tool for students studying thermodynamics.

isobaric process

An isobaric process is one where the pressure of the gas remains constant throughout the transformation. During this process, any heat added to the system does work and causes a change in the volume of the gas. When plotted on a PV diagram, the process will appear as a horizontal line.

For the problem in question, the volume of the gas doubled while keeping the pressure constant at 2.40 \times 10^5 Pa. To calculate the work done during an isobaric process, use the equation \(W = P\Delta V\), where \(\Delta V\) is the change in volume. Unlike isochoric processes where no work occurs, isobaric processes do involve work and energy transfer, which is often a key point to grasp in thermodynamics.

Understanding this process helps in analyzing engines and machinery that operate under constant pressures, making it a fundamental part of engineering applications.

isothermal process

An isothermal process occurs when a gas changes state or shape at a constant temperature. During an isothermal process, energy absorbed by the system as heat is exactly matched by the work done by the system, so the temperature does not change.

In the exercise, the gas was compressed back to its original volume while maintaining a temperature of 355 K, resulting in an isothermal curve on the PV diagram. For an ideal gas, the equation governing this process is \(PV = nRT\). The work done in isothermal compression can be calculated using \(W = nRT \ln\left(\frac{V_f}{V_i}\right)\), which accounts for the logarithmic relationship involved when compressing the gas.

This concept is crucial for understanding how refrigerators and air conditioners work because they often utilize the thermodynamic principles of isothermal processes to maintain temperature constancy.

isochoric process

An isochoric process occurs when a gas undergoes changes without any variation in its volume. In simpler terms, this means that the gas is confined to a space that prevents it from expanding or contracting. Therefore, all the heat added or removed from the system changes the pressure and temperature, but not the volume.

In terms of the PV diagram in the exercise, the isochoric process is depicted by a vertical line, indicating constant volume. As such, no work is done, because work involves a change in volume (as \(W = P\Delta V\) implies no volume change \(\Delta V = 0\)). This process concludes the thermodynamic cycle described in the exercise, as it returns the system to its original pressure.

Understanding this concept is significant in areas like calorimetry, where changes in temperature at constant volume help determine heat capacities and other vital thermal properties.